NYT Pips Hints & Answers for April 11, 2026

Apr 11, 2026

🚨 SPOILER WARNING

This page contains the final **answer** and the complete **solution** to today's NYT Pips puzzle. If you haven't attempted the puzzle yet and want to try solving it yourself first, now's your chance!

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🎲 Today's Puzzle Overview

Ian Livengood's easy grid for April 11th opens at the top and works its way down. A single-cell sum=5 constraint at the top of the board names the only domino that can reach there with a 5-pip face pointing upward, and its partner drops into an empty cell just below. The two three-cell sum regions that run down the board's two columns are each resolved by a double and a non-double working together — the [4|4] double anchors the left column, and the only domino with a 0-pip face settles the right. A lone single-cell sum=5 at the corner confirms the last placement and closes the puzzle.

Rodolfo Kurchan's medium puzzle for April 11th is a double-heavy board: four of the seven dominoes are doubles — [0|0], [4|4], [5|5], and [6|6] — and each one occupies a distinct region. The three-cell equals region in the center row is the anchor point: any double placed inside it forces all three cells to the same value. The [6|6] double is the only tile that can fill two adjacent cells in that region while still allowing the surrounding sum=6 and sum=10 constraints to resolve. With that one placement, two sum=10 regions across the top and bottom close in sequence, and two more doubles fall into their slots without ambiguity.

Rodolfo Kurchan's hard puzzle for April 11th is ruled by a sum=1 region in the top row — a two-cell constraint so tight that only a single pair of pip values (0 and 1) can satisfy it. That placement immediately fixes values in four adjacent cells through a chain of sums, locking down the entire top row in one pass. Below the top row, a three-cell equals region of 2s is anchored by the only double in today's hard set. A four-cell less-than-5 region at the bottom is filled entirely with 1-pip faces — three supplied by separate non-double dominoes, and one by the [1|1] double that closes the board.

💡 Progressive Hints

Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!

💡 Sum=5 at the top names the first domino
The single cell at the top of the grid must show exactly 5 pips. Only one domino in today's easy set has a 5-pip face available to reach that position. Place it, and its partner drops into the empty cell just below — no constraint to satisfy there.
💡 A double anchors the left column; a 0-pip domino anchors the right
The three-cell sum=9 in the left column and sum=7 in the right column each need two more dominoes. The [4|4] double fills two cells of the left column with 4s, leaving the remaining cell already determined. For the right column, one domino carries a 0-pip face — that zero anchors the sum=7 check, and the partner face gives you the last single-cell sum=5 at the corner.
💡 Full solution
Sum=5 at (0,1): [2|5] covers (1,1) and (0,1) — 2 at (1,1) (empty ✓), 5 at (0,1) ✓. Sum=9 at (2,0)+(3,0)+(4,0): [0|1] covers (2,1) and (2,0) — 0 at (2,1) (empty ✓), 1 at (2,0). [4|4] double covers (3,0) and (4,0) — 4 at each. Sum: 1+4+4=9 ✓. Sum=7 at (2,2)+(3,2)+(4,2): [3|5] covers (2,2) and (2,3) — 3 at (2,2), 5 at (2,3). Sum=5 at (2,3)=5 ✓. [0|4] covers (3,2) and (4,2) — 0 at (3,2), 4 at (4,2). Sum: 3+0+4=7 ✓. All six constraints satisfied. Puzzle complete.
💡 The three-cell equals region in the center hosts the [6|6] double
Four doubles sit in today's medium set. The three-cell equals region at row 2 requires all three cells to share one pip value. The [6|6] double placed horizontally inside that region pins two adjacent cells to 6 immediately, forcing the third to match. No other double can claim this region cleanly given the two sum=10 constraints flanking the board.
💡 Both sum=10 regions close from the equals anchor
With row 2 locked at 6, the sum=6 region above it gives you one cell's value, and both sum=10 regions — one across the top and one at the bottom — can be read off in sequence. The remaining doubles each slot into an unconstrained pair: [5|5] satisfies sum=10 horizontally, [4|4] satisfies sum=10 at the bottom, and [0|0] fills the top-left empties.
💡 Full solution
Three-cell equals (2,0),(2,1),(2,2): [6|6] double covers (2,0) and (2,1) horizontally — 6+6. (2,2) must also be 6. [3|6] covers (1,2) and (2,2): 3 at (1,2), 6 at (2,2) ✓. Sum=6 at (0,2)+(1,2): (1,2)=3, so (0,2)=3. [3|1] covers (0,2) and (0,1): 3 at (0,2) ✓, 1 at (0,1) (empty ✓). Sum=10 at (1,3)+(1,4)=10: [5|5] double covers (1,3) and (1,4) — 5+5=10 ✓. Sum=10 at (3,3)+(3,4)=10: [4|4] double covers (3,2) and (3,3) — 4 at (3,2) (empty ✓), 4 at (3,3). (3,4) must be 6. [2|6] covers (2,4) and (3,4): 2 at (2,4), 6 at (3,4) ✓. Less-than-4 at (2,4)=2 ✓. [0|0] double covers (0,0) and (1,0): 0 at each (all empty ✓). All nine constraints satisfied. Puzzle complete.
💡 Sum=1 across two cells — the board's tightest constraint
The two-cell region at (0,1) and (0,2) must total exactly 1. Only two pip combinations produce 1: 0+1 or 1+0. Scan today's hard set for a domino that can split a 0-pip and a 1-pip face across those two cells. There is exactly one arrangement. Place it, and the values cascade into the adjacent sum=5 and sum=10 regions — locking the entire top row in one pass.
💡 The [2|2] double anchors the three-cell equals region of 2s
Below the top row, a three-cell equals region spanning the left side of the board requires all three cells to match. The [2|2] double is today's only double in the hard set, and it drops directly into two of those three cells. The third cell is already fixed from the top-row cascade, and it confirms the shared value of 2. A chain of less-than and greater-than constraints then fills the remaining board without backtracking.
💡 Full solution
Sum=1 at (0,1)+(0,2)=1: [2|1] covers (0,0) and (0,1) — 2 at (0,0), 1 at (0,1). [0|5] covers (0,2) and (0,3) — 0 at (0,2), 5 at (0,3). 1+0=1 ✓. Sum=10 at (0,3)+(0,4): (0,3)=5, so (0,4)=5. [4|5] covers (1,4) and (0,4): 4 at (1,4) (empty ✓), 5 at (0,4) ✓. Sum=5 at (0,0)+(1,0): (0,0)=2, so (1,0)=3. [3|6] covers (1,0) and (2,0): 3 at (1,0) ✓, 6 at (2,0). Greater-than-5 at (2,0)=6 ✓. Less-than-5 at (2,2)+(2,3): [0|4] covers (2,2) and (2,3): 0 at (2,2), 4 at (2,3) — 0+4=4<5 ✓. Equals (3,0),(4,0),(4,1): [2|2] double covers (3,0) and (4,0): 2 at (3,0), 2 at (4,0). [2|3] covers (4,1) and (4,2): 2 at (4,1) ✓, 3 at (4,2). Less-than-5 at (4,2)=3 ✓. Greater-than-5 at (4,4)=6: [1|6] covers (5,4) and (4,4): 1 at (5,4), 6 at (4,4) ✓. Sum=5 at (5,0)=5: [2|5] covers (6,0) and (5,0): 2 at (6,0), 5 at (5,0) ✓. Sum=5 at (6,0)+(6,1): (6,0)=2, so (6,1)=3. [1|3] covers (6,2) and (6,1): 1 at (6,2), 3 at (6,1) ✓. Less-than-5 region (5,4),(6,2),(6,3),(6,4): (5,4)=1 ✓, (6,2)=1 ✓. [1|1] double covers (6,3) and (6,4): 1 at (6,3) ✓, 1 at (6,4) ✓. [4|1] covers (2,4) and (3,4): 4 at (2,4) (empty ✓), 1 at (3,4) (empty ✓). All fourteen constraints satisfied. Puzzle complete.

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Apr 11, 2026

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Final Answer & Complete Solution For Hard Level

The key to solving today's hard puzzle was identifying the placement for the critical dominoes highlighted in the starting grid. Once those were in place, the rest of the puzzle could be solved logically. See the final grid below to compare your solution.

Starting Position & Key First Steps

Pips hint for April 11, 2026 – hard level puzzle grid with critical first placements and strategy

This image shows the initial puzzle grid for the hard level, with a few critical first placements highlighted.

Final Answer: The Solved Grid for Hard Mode

NYT Pips April 11, 2026 hard puzzle full solution grid showing final answer with hints

Compare this final grid with your own solution to see the correct placement of all dominoes.

🔧 Step-by-Step Answer Walkthrough For Easy Level

1
Step 1: Sum=5 at (0,1) points to a single domino
Cell (0,1) must show exactly 5 pips. Today's easy set has two dominoes with a 5-pip face: [2|5] and [3|5]. Of these, only [2|5] can orient its 5-pip face into (0,1) from the adjacent cell (1,1). Place [2|5] vertically: 2 at (1,1) — empty constraint ✓ — and 5 at (0,1) ✓.
2
Step 2: Sum=9 in the left column — the double and the 1-pip
The region at (2,0),(3,0),(4,0) must total 9. The [4|4] double covers (3,0) and (4,0): 4+4=8. The remaining cell (2,0) must contribute 1. The [0|1] domino covers (2,1) and (2,0): 0 at (2,1) — empty ✓ — and 1 at (2,0) ✓. Sum: 1+4+4=9 ✓.
3
Step 3: Sum=5 at (2,3) names the next domino
Cell (2,3) must show exactly 5. The [3|5] domino covers (2,2) and (2,3): 3 at (2,2), 5 at (2,3) ✓.
4
Step 4: Sum=7 in the right column confirms the final placement
The region at (2,2),(3,2),(4,2) must total 7. Cell (2,2)=3 is now fixed. The [0|4] domino covers (3,2) and (4,2): 0 at (3,2), 4 at (4,2). Sum: 3+0+4=7 ✓. All six constraints satisfied. Puzzle complete.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: [6|6] double anchors the three-cell equals region
The equals region at (2,0),(2,1),(2,2) requires all three cells to share one pip value. The [6|6] double placed horizontally at (2,0) and (2,1) pins both to 6, forcing (2,2)=6 as well. Three cells resolved in one placement.
2
Step 2: Sum=6 at (0,2)+(1,2) and the top row
Cell (2,2)=6 is fixed. The [3|6] domino covers (1,2) and (2,2): 3 at (1,2), 6 at (2,2) ✓. Sum=6 at (0,2)+(1,2): (1,2)=3, so (0,2)=3. [3|1] covers (0,2) and (0,1): 3 at (0,2) ✓, 1 at (0,1) — empty ✓.
3
Step 3: [5|5] double fills sum=10 at the top
Sum=10 at (1,3)+(1,4)=10: the [5|5] double placed horizontally — 5+5=10 ✓. No other single domino in today's set reaches 10.
4
Step 4: [4|4] double fills sum=10 at the bottom
Sum=10 at (3,3)+(3,4)=10: [4|4] placed with 4 at (3,2) — empty ✓ — and 4 at (3,3). So (3,4) must be 6. [2|6] covers (2,4) and (3,4): 2 at (2,4), 6 at (3,4) ✓. Less-than-4 at (2,4)=2 ✓.
5
Step 5: [0|0] double fills the top-left empties
The three remaining empty cells (0,0) and (1,0) accept the [0|0] double — 0 at (0,0) and 0 at (1,0). All empty constraints satisfied ✓. All nine constraints across the board satisfied. Puzzle complete.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Sum=1 at (0,1)+(0,2) — only one possible split
A total of 1 from two cells means one holds pip 0 and the other holds pip 1. The [2|1] domino covers (0,0) and (0,1): 2 at (0,0), 1 at (0,1). The [0|5] domino covers (0,2) and (0,3): 0 at (0,2), 5 at (0,3). Sum: 1+0=1 ✓.
2
Step 2: Sum=10 and sum=5 close the top row
Sum=10 at (0,3)+(0,4): (0,3)=5, so (0,4)=5. [4|5] covers (1,4) and (0,4): 4 at (1,4) — empty ✓ — and 5 at (0,4) ✓. Sum=5 at (0,0)+(1,0): (0,0)=2, so (1,0)=3. [3|6] covers (1,0) and (2,0): 3 at (1,0) ✓, 6 at (2,0). Greater-than-5 at (2,0)=6 ✓. Top row fully resolved.
3
Step 3: Less-than-5 across row 2
The region at (2,2) and (2,3) must total less than 5. [0|4] covers (2,2) and (2,3): 0 at (2,2), 4 at (2,3) — 0+4=4<5 ✓. The remaining two cells in row 2 carry empty constraints: [4|1] covers (2,4) and (3,4): 4 at (2,4) ✓, 1 at (3,4) ✓.
4
Step 4: [2|2] double anchors the three-cell equals region
The equals region at (3,0),(4,0),(4,1) requires all three to match. [2|2] double covers (3,0) and (4,0): 2 at each. So (4,1) must also be 2. [2|3] covers (4,1) and (4,2): 2 at (4,1) ✓, 3 at (4,2). Less-than-5 at (4,2)=3 ✓.
5
Step 5: Greater-than-5 at (4,4) and sum=5 at (5,0)
Greater-than-5 at (4,4): must be 6. [1|6] covers (5,4) and (4,4): 1 at (5,4), 6 at (4,4) ✓. Sum=5 at (5,0)=5: [2|5] covers (6,0) and (5,0): 2 at (6,0), 5 at (5,0) ✓.
6
Step 6: Sum=5 at (6,0)+(6,1) and the four-cell less-than region
Sum=5 at (6,0)+(6,1): (6,0)=2, so (6,1)=3. [1|3] covers (6,2) and (6,1): 1 at (6,2) ✓, 3 at (6,1) ✓. Less-than-5 region (5,4),(6,2),(6,3),(6,4): (5,4)=1<5 ✓, (6,2)=1<5 ✓. [1|1] double covers (6,3) and (6,4): 1 at each — both <5 ✓. All fourteen constraints satisfied. Puzzle complete.

💡 Pro Tips for Similar Puzzles

Start with Constraints
Always begin with the most constrained regions - sum regions with small numbers or tight spaces.
Use Equal Regions
Use "equal" regions as anchors - they eliminate many possibilities quickly.
Work Systematically
Let the rules guide your placement rather than guessing randomly.
Double-Check
Verify each region's rules are satisfied before moving to the next.

🎓 Keep Learning & Improve